The proof consists of truncation of \(\rho =\rho _{u,v}\)  given by [1]} and the lemma above obtained using duality.
(Step 1)
Let us assume that \(\Sigma \)  is invertible. Otherwise, we only need to consider a lower-dimensional subspace, and the proof is indifferent to the case with invertible \(\Sigma \) .
Let \(\mu \)  denote a \(2p\) -dimensional product measure of two \(p\) -dimensional Gaussian measures with a zero mean and convariance \(\left(2\mathbf {r}\left(\Sigma \right)\right)^{-1}\Sigma \) .
We define \(\mathbb {S}^{p-1}\)  as a unit ball in \(\mathbb {R}^{p}\) .
Let us set \(u,v\in \Sigma ^{1/2}\mathbb {S}^{p-1}\)  and define \(f_{u},f_{v}\)  as probability density functions with respect to the Lebesgue measure such that
\(f_{u}\left(x\right)&=\frac{\exp \left(-\mathbf {r}\left(\Sigma \right)\left(x-u\right)^{\top }\Sigma ^{-1}\left(x-u\right)\right)\mathbf {1}\lbrace \left\Vert x-u\right\Vert \le \sqrt{\left\Vert \Sigma \right\Vert }\rbrace }{ \int \exp \left(-\mathbf {r}\left(\Sigma \right)\left(x^{\prime }-u\right)^{\top }\Sigma ^{-1}\left(x^{\prime }-u\right)\right)\mathbf {1}\lbrace \left\Vert x^{\prime }-u\right\Vert \le \sqrt{\left\Vert \Sigma \right\Vert }\rbrace  \mathrm {d}x^{\prime }},\\f_{v}\left(x\right)&=\frac{\exp \left(-\mathbf {r}\left(\Sigma \right)\left(x-u\right)^{\top }\Sigma ^{-1}\left(x-v\right)\right)\mathbf {1}\lbrace \left\Vert x-v\right\Vert \le \sqrt{\left\Vert \Sigma \right\Vert }\rbrace }{ \int \exp \left(-\mathbf {r}\left(\Sigma \right)\left(x^{\prime }-v\right)^{\top }\Sigma ^{-1}\left(x^{\prime }-v\right)\right)\mathbf {1}\lbrace \left\Vert x^{\prime }-v\right\Vert \le \sqrt{\left\Vert \Sigma \right\Vert }\rbrace  \mathrm {d}x^{\prime }}.\) 

Here, \(\mathbf {1}\lbrace \mathcal {E}\rbrace \)  is an indicator function which is 1 if an event \(\mathcal {E}\)  is true and 0 otherwise.
Assume that the independent random vectors \(\theta ,\eta \)  have densities \(f_{u}\)  and \(f_{v}\) , then \(\mathbb {E}[\left(\theta ,\eta \right)]=\left(u,v\right)\)  by the symmetricity of \(f_{u},f_{v}\) , and \(\max \left\lbrace \left\Vert \theta \right\Vert ,\left\Vert \eta \right\Vert \right\rbrace \le 2\sqrt{\left\Vert \Sigma \right\Vert }\)  almost surely.
Let \(\rho _{u,v}\)  be a probability measure of \(\theta ,\eta \)  given by \(\rho _{u,v}\left(\mathrm {d}x,\mathrm {d}y\right)=f_{u}\left(x\right)f_{v}\left(y\right)\mathrm {d}x\mathrm {d}y,x,y\in \mathbb {R}^{d}\) .
In the proof of Theorem 1 of [1]},
it is shown that
\(\mathrm {KL}\left(\rho _{u,v} \Vert  \mu \right)\le 2\log 2+2\mathbf {r}\left(\Sigma \right).\) 
(Step 2)
Let \(f\left(A,\theta ,\eta \right):=\theta ^{\top }A\eta \)  for any \(A\in \mathbb {R}^{p}\otimes \mathbb {R}^{p}\)  and \(\theta ,\eta \in \mathbb {R}^{p}\) .
Lemma REF  with \(h(M_1,\dots ,M_n) = \sum _{\ell =1}^n f(M_\ell ,\theta ,\eta )\)  gives that for any \(\lambda >0\) ,
\(&\mathbb {E}_{\mathbf {M}}\left[\exp \left(\lambda \sum _{\ell =1}^{n}f\left(M_{\ell },\theta ,\eta \right)\right)\right]\\&=\mathbb {E}_{\mathbf {M}}\left[\exp \left(\lambda \sum _{\ell =1}^{n}\theta ^{\top }M_{\ell }\eta \right)\right]\\&\le \exp \left(n\lambda \theta ^{\top }\Sigma \eta +\frac{n^{2}\lambda ^{2}\left\Vert \theta \right\Vert ^{2}\left\Vert \eta \right\Vert ^{2}\left(2\kappa ^{2}+\Gamma _{n}\right)^{2}}{8n}\right),\) 

because for any \(A_{1},\ldots ,A_{n},B_{1},\ldots ,B_{n}\in \mathbb {R}^{p}\otimes \mathbb {R}^{p}\) ,
\(\left|h\left(A_{1},\dots ,A_{n}\right)-h\left(B_{1},\dots ,B_{n}\right)\right|&=\left|\theta ^{\top }\left(\sum _{\ell =1}^{n}\left(A_{\ell }-B_{\ell }\right)\right)\eta \right|\\&\le \left\Vert \theta \right\Vert \left\Vert \eta \right\Vert \sum _{\ell =1}^{n}\left\Vert A_{\ell }-B_{\ell }\right\Vert .\) 

It holds that
\(&\frac{1}{n}\mathbb {E}_{\rho _{u,v}}\left[\log \mathbb {E}_{\mathbf {M}}\left[\exp \left(\lambda \sum _{\ell =1}^{n}f\left(M_{\ell },\theta ,\eta \right)\right)\right]\right]\\&\le \frac{1}{n}\mathbb {E}_{\rho _{u,v}}\left[n\lambda \theta ^{\top }\Sigma \eta +\frac{n^{2}\lambda ^{2}\left\Vert \theta \right\Vert ^{2}\left\Vert \eta \right\Vert ^{2}\left(2\kappa ^{2}+\Gamma _{n}\right)^{2}}{8n}\right]\\&=\mathbb {E}_{\rho _{u,v}}\left[\lambda \theta ^{\top }\Sigma \eta +\frac{\lambda ^{2}\left\Vert \theta \right\Vert ^{2}\left\Vert \eta \right\Vert ^{2}\left(2\kappa ^{2}+\Gamma _{n}\right)^{2}}{8}\right]\\&\le \lambda u^{\top }\Sigma v+\frac{\lambda ^{2}\left(2\sqrt{\left\Vert \Sigma \right\Vert }\right)^{4}\left(2\kappa ^{2}+\Gamma _{n}\right)^{2}}{8}\\&=\lambda u^{\top }\Sigma v+2\lambda ^{2}\left\Vert \Sigma \right\Vert ^{2}\left(2\kappa ^{2}+\Gamma _{n}\right)^{2}.\) 

The last inequality follows the fact \(\max \left\lbrace \left\Vert \theta \right\Vert ,\left\Vert \eta \right\Vert \right\rbrace \le 2\sqrt{\left\Vert \Sigma \right\Vert }\) .
Therefore, from Lemma REF  with \(h\left(M_{1},\ldots ,M_{n},\theta ,\eta \right)=\lambda \sum _{\ell =1}^{n}f\left(M_{\ell },\theta ,\eta \right)\)  and the fact that \(\log 2\le \mathbf {r}\left(\Sigma \right)\)  for any \(\Sigma \) , we obtain
\(\frac{1}{n}\sum _{\ell =1}^{n}\lambda u^{\top }M_{\ell }v\le \lambda u^{\top }\Sigma v+2\lambda ^{2}\left\Vert \Sigma \right\Vert ^{2}\left(2\kappa ^{2}+\Gamma _{n}\right)^{2}+\frac{4\mathbf {r}\left(\Sigma \right)+t}{n},\) 

simultaneously for all \(u,v\)  with probability at least \(1-e^{-t}\) .
By choosing
\(\lambda =\sqrt{\frac{4\mathbf {r}\left(\Sigma \right)+t}{2n\left\Vert \Sigma \right\Vert ^{2}\left(2\kappa ^{2}+\Gamma _{n}\right)^{2}}},\) 

we obtain a bound such that
\(\left\Vert \frac{1}{n}\sum _{\ell =1}^{n} M_{\ell }-\Sigma \right\Vert \le 2\sqrt{2}\left\Vert \Sigma \right\Vert \left(2\kappa ^{2}+\Gamma _{n}\right)\sqrt{\frac{4\mathbf {r}\left(\Sigma \right)+t}{n}}.\) 

This is our claim.
